(a) In the standard notation \xi \equiv \sqrt{m \omega / \hbar} x, \alpha \equiv(m \omega / \pi \hbar)^{1 / 4} .
\Psi(x, 0)=A(1-2 \xi)^{2} e^{-\xi^{2} / 2}=A\left(1-4 \xi+4 \xi^{2}\right) e^{-\xi^{2} / 2} .
It can be expressed as a linear combination of the first three stationary states (Eq. 2.60 and 2.63, and Problem 2.10):
\psi_{0}(x)=\left(\frac{m \omega}{\pi \hbar}\right)^{1 / 4} e^{-\frac{m \omega}{2 \hbar} x^{2}} (2.60).
\psi_{1}(x)=A_{1} \hat{a}_{+} \psi_{0}=\frac{A_{1}}{\sqrt{2 \hbar m \omega}}\left(-\hbar \frac{d}{d x}+m \omega x\right)\left(\frac{m \omega}{\pi \hbar}\right)^{1 / 4} e^{-\frac{m \omega}{2 \hbar} x^{2}}
=A_{1}\left(\frac{m \omega}{\pi \hbar}\right)^{1 / 4} \sqrt{\frac{2 m \omega}{\hbar}} x e^{-\frac{m \omega}{2 \hbar} x^{2}} (2.63).
\psi_{0}(x)=\alpha e^{-\xi^{2} / 2} ,
\psi_{1}(x)=\sqrt{2} \alpha \xi e^{-\xi^{2} / 2} ,
\psi_{2}(x)=\frac{\alpha}{\sqrt{2}}\left(2 \xi^{2}-1\right) e^{-\xi^{2} / 2} .
So \Psi(x, 0)=c_{0} \psi_{0}+c_{1} \psi_{1}+c_{2} \psi_{2}=\alpha\left(c_{0}+\sqrt{2} \xi c_{1}+\sqrt{2} \xi^{2} c_{2}-\frac{1}{\sqrt{2}} c_{2}\right) e^{-\xi^{2} / 2} , with (equating like powers)
\begin{cases}\alpha \sqrt{2} c_{2}=4 A & \Rightarrow c_{2}=2 \sqrt{2} A / \alpha \\ \alpha \sqrt{2} c_{1}=-4 A & \Rightarrow c_{1}=-2 \sqrt{2} A / \alpha \\ \alpha\left(c_{0}-c_{2} / \sqrt{2}\right)=A & \Rightarrow c_{0}=(A / \alpha)+c_{2} / \sqrt{2}=(1+2) A / \alpha=3 A / \alpha\end{cases} .
Normalizing: 1=\left|c_{0}\right|^{2}+\left|c_{1}\right|^{2}+\left|c_{2}\right|^{2}=(8+8+9)(A / \alpha)^{2}=25(A / \alpha)^{2} \Rightarrow A=\alpha / 5 = \frac{1}{5}\left(\frac{m \omega}{\pi \hbar}\right)^{1 / 4} .
c_{0}=\frac{3}{5}, \quad c_{1}=-\frac{2 \sqrt{2}}{5}, \quad c_{2}=\frac{2 \sqrt{2}}{5} .
(b) You could get \frac{1}{2} \hbar \omega probability, \frac{9}{25} ; \frac{3}{2} \hbar \omega, , probability \frac{8}{25} ; \frac{5}{2} \hbar \omega, , probability \frac{8}{25}
\langle H\rangle=\frac{9}{25}\left(\frac{1}{2} \hbar \omega\right)+\frac{8}{25}\left(\frac{3}{2} \hbar \omega\right)+\frac{8}{25}\left(\frac{5}{2} \hbar \omega\right)=\frac{\hbar \omega}{50}(9+24+40) = \frac{73}{50} \hbar \omega .
(c)
\Psi(x, t)=\frac{3}{5} \psi_{0} e^{-i \omega t / 2}-\frac{2 \sqrt{2}}{5} \psi_{1} e^{-3 i \omega t / 2}+\frac{2 \sqrt{2}}{5} \psi_{2} e^{-5 i \omega t / 2}=e^{-i \omega t / 2}\left[\frac{3}{5} \psi_{0}-\frac{2 \sqrt{2}}{5} \psi_{1} e^{-i \omega t}+\frac{2 \sqrt{2}}{5} \psi_{2} e^{-2 i \omega t}\right] .
To change the sign of the middle term we need e^{-i \omega T}=-1 (then e^{-2 i \omega T}=1 ); evidently \omega T=\pi , or T = π/ω.